$\alpha = \frac{2}{3+\sqrt{5}}$ と $\beta = \frac{2}{3-\sqrt{5}}$ が与えられたとき、以下の値を求める問題です。 (1) $\alpha^2 + \beta^2$ (2) $\frac{\beta}{\alpha-1} + \frac{\alpha}{\beta-1}$ (3) $\sqrt{\alpha} + \sqrt{\beta}$

1. 問題の内容

$\alpha = \frac{2}{3+\sqrt{5}}$ と $\beta = \frac{2}{3-\sqrt{5}}$ が与えられたとき、以下の値を求める問題です。

(1) $\alpha^2 + \beta^2$

(2) $\frac{\beta}{\alpha-1} + \frac{\alpha}{\beta-1}$

(3) $\sqrt{\alpha} + \sqrt{\beta}$

2. 解き方の手順

まず、$\alpha$ と $\beta$ を簡単にします。

$\alpha = \frac{2}{3+\sqrt{5}} = \frac{2(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})} = \frac{2(3-\sqrt{5})}{9-5} = \frac{2(3-\sqrt{5})}{4} = \frac{3-\sqrt{5}}{2}$

$\beta = \frac{2}{3-\sqrt{5}} = \frac{2(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})} = \frac{2(3+\sqrt{5})}{9-5} = \frac{2(3+\sqrt{5})}{4} = \frac{3+\sqrt{5}}{2}$

(1) $\alpha^2 + \beta^2$ を求めます。

$\alpha^2 = (\frac{3-\sqrt{5}}{2})^2 = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2}$

$\beta^2 = (\frac{3+\sqrt{5}}{2})^2 = \frac{9 + 6\sqrt{5} + 5}{4} = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2}$

$\alpha^2 + \beta^2 = \frac{7 - 3\sqrt{5}}{2} + \frac{7 + 3\sqrt{5}}{2} = \frac{7 - 3\sqrt{5} + 7 + 3\sqrt{5}}{2} = \frac{14}{2} = 7$

(2) $\frac{\beta}{\alpha-1} + \frac{\alpha}{\beta-1}$ を求めます。

$\alpha - 1 = \frac{3-\sqrt{5}}{2} - 1 = \frac{3-\sqrt{5}-2}{2} = \frac{1-\sqrt{5}}{2}$

$\beta - 1 = \frac{3+\sqrt{5}}{2} - 1 = \frac{3+\sqrt{5}-2}{2} = \frac{1+\sqrt{5}}{2}$

$\frac{\beta}{\alpha-1} = \frac{\frac{3+\sqrt{5}}{2}}{\frac{1-\sqrt{5}}{2}} = \frac{3+\sqrt{5}}{1-\sqrt{5}} = \frac{(3+\sqrt{5})(1+\sqrt{5})}{(1-\sqrt{5})(1+\sqrt{5})} = \frac{3 + 3\sqrt{5} + \sqrt{5} + 5}{1-5} = \frac{8 + 4\sqrt{5}}{-4} = -2 - \sqrt{5}$

$\frac{\alpha}{\beta-1} = \frac{\frac{3-\sqrt{5}}{2}}{\frac{1+\sqrt{5}}{2}} = \frac{3-\sqrt{5}}{1+\sqrt{5}} = \frac{(3-\sqrt{5})(1-\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})} = \frac{3 - 3\sqrt{5} - \sqrt{5} + 5}{1-5} = \frac{8 - 4\sqrt{5}}{-4} = -2 + \sqrt{5}$

$\frac{\beta}{\alpha-1} + \frac{\alpha}{\beta-1} = -2 - \sqrt{5} + (-2 + \sqrt{5}) = -4$

(3) $\sqrt{\alpha} + \sqrt{\beta}$ を求めます。

$\alpha = \frac{3-\sqrt{5}}{2} = \frac{6-2\sqrt{5}}{4} = \frac{5 - 2\sqrt{5} + 1}{4} = \frac{(\sqrt{5}-1)^2}{4}$

$\beta = \frac{3+\sqrt{5}}{2} = \frac{6+2\sqrt{5}}{4} = \frac{5 + 2\sqrt{5} + 1}{4} = \frac{(\sqrt{5}+1)^2}{4}$

$\sqrt{\alpha} = \sqrt{\frac{(\sqrt{5}-1)^2}{4}} = \frac{\sqrt{5}-1}{2}$

$\sqrt{\beta} = \sqrt{\frac{(\sqrt{5}+1)^2}{4}} = \frac{\sqrt{5}+1}{2}$

$\sqrt{\alpha} + \sqrt{\beta} = \frac{\sqrt{5}-1}{2} + \frac{\sqrt{5}+1}{2} = \frac{\sqrt{5}-1+\sqrt{5}+1}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$

3. 最終的な答え

(1) 7

(2) -4

(3) $\sqrt{5}$